Recent content by Eighty

f(x)=sin(1/x)/x.

You won't get a 0 in the denominator. You forgot to divide the 1 by n^2.

f is the function, f(n) is the value of the function at the point n. The ball operator \circ acts on functions, not values, so your second expression is wrong.

Depends on the book/subject. function, map, transform and operator are usually synonyms, but some books define one of the latter to be something special, if it's used a lot. E.g. a map may be defined to be a continuous function, or a homeomorphism. Transforms and operators are usually linear...

Depends on what you mean by related. The method of showing denseness and countability will probably be very similar, so much in fact that you'd be inclined to say it's the same thing. Another countable dense set of the reals is the set of all integer multiples of a sequence of reals...

No, you can take that sequence, and remove a finite number of elements or add infinitely many elements and it will have the same property. A denumeration of the set of rationals of the form n/2^k for all n and k\geq 0 (ie all binary numbers with a finite number of digits) also works.

Assuming f and g_n are complex-valued, iff f is continuous.

Choose an appropriate x.

A function f(x) has an asymptote if it "looks like" a straight line when x tends to \pm\infty. More precisely, if the distance |f(x)-\text{the line}| tends to 0. A parabola doesn't have one. Suppose there was one (draw an arbitrary line in the same graph). Then, at some point, the parabola...

Both your definition and proof are correct.

Yes, when the third implies the second. Ie a\Rightarrow b\Rightarrow a gives you a\Leftrightarrow b. The chain of implications really is: 1/x=1/(2x+1) \Rightarrow x=2x+1 \Leftrightarrow x=-1 \Rightarrow 1/x=1/(2x+1) Since 1/x=1/(2x+1) \Rightarrow x=2x+1 and x=2x+1\Rightarrow 1/x=1/(2x+1), we...

You know that b is the least upper bound. So b-1, for example, isn't an upper bound. Neither is b-1/2, or b-1/4, etc... b-epsilon will not be an upper bound for any choice of epsilon>0. Now, since b-epsilon is not an upper bound, there must be an s in S such that b-epsilon<=s, or else...

a=b implies 1/a=1/b iff a=b\neq 0. In this case, x can't be 0, so it won't be the case, but maybe you didn't know that until after that particular step. Once you know that x isn't 0, you can put an equivalence arrow there, but in general you cannot.

These almost-counter examples to Fermat's theorem were in Simpsons: (1782^{12} + 1841^{12})^{1/12}\approx1921.99999995587 (3987^{12} + 4365^{12})^{1/12}\approx4472.00000000706

1^i = 1. 1^x = 1 for all complex numbers x. The complex numbers are closed, so you'll never need another unit j, say, to solve equations involving complex numbers, such as 1^i=x. Lengths are positive real numbers. You can't have something with an imaginary length. Putting complex...